Hello.
$$ \textbf{Going back to the linear ODE from its transfer function} $$
The transfer function here:
$$ G(s) = \dfrac{2s + 1}{ s^{2} + 2s + 5} $$
Use the laplace transform properties of differentiation, and the definition of the transfer function itself to do the above step. The transfer function is the ratio of the output over input when the input is the dirac pulse.
This describes the linear ODE:
$$ 2 x'(t) + x(t) = y''(t) + 2y'(t) + 5y(t) \tag{System} $$
Solve for this ODES homogenous response, setting $$ x(t) = 0 $$
$$ y'' + 2y' + 5y(t) = 0 $$
Using the standard method for solving such ODES, the quadratic equation and the roots, yields a complex solution: $$ z_{1, 2} = -1 \pm 2j $$ Our homogenous solution would then be: $$ y_{h}(t) = e^{-t} \cdot \Big ( A_{1} \cos(2t) + B_{1} \sin(2t) \Big) \tag{Damped sinusoidal excitation} $$
This can be expressed in one sinusoid (using addition by complex numbers):
$$ y_{h}(t) = e^{-t} \cdot \Big ( {\sqrt{A_{1}^{2} + B_{1}^{2} } } \cdot \cos(2t - \phi_{1} )) \Big) \,\,\,\,A_{1}, B_{1} \in \mathbb{R} \tag{Damped sinusoidal excitation} $$
Where:
$$ \phi_{1} = \text{Arg}\Big[ A_{1} - j B_{1} \Big] $$
So, the damping ratio would be -1 the natural frequency:
$$ 2 \,\,\,\, \dfrac{ \text{Rad} }{ \text{Sec} } $$
You can take the inverse laplace transform of G(s) and see for yourself:
$$
G(s) \rightarrow g(t) = 2e^{-t} \cos(2t) - \dfrac{1}{2} e^{-t} \sin(t)
$$
$$
g(t) = y_{h}(t)= \dfrac{\sqrt{17}e^{-t} }{2} \cos(2t - \arctan(0.25) )
$$
$$ \text{The inverse laplace transform of a systems transfer function is the homogenous solution of that systems ODE} $$
$$ \textbf{My questions, please see } $$
I am not 100 percent sure if my solution is correct. Can you please confirm? If possible, later you can suggest an alternative path too. secondly why is the transfer function denoted G(s) ? It is usually denoted H(s) (impulse response) where G(s) is the unit step response.